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Answers to Study and Review Questions Answers to Study and Review Questions Answers are given below for calculations, problems, and short 4. It is an accident in the sense that other codes with the same answers; a reference to the relevant section(s) of the chapter is four letters could work equally well. It is frozen in the sense given for longer answers and definitions or explanations of that changes in it are selected against. See Section 3.8. technical terms; or a reference to further reading is given for 5. This is intended more as a discussion topic: for the idea see topics not explicitly discussed in the text. Section 3.9. 6. Because a form would have existed in time before the series Chapter 1 of fossils (of vertebrates from fish to mammals) that can be 1. See Section 1.1. strongly argued to be its ancestors. 2. Adaptation. 3. The popular concept had evolution as progressive, with Chapter 4 species ascending a one-dimensional line from lower 1. forms to higher. Evolution in Darwin’s theory is tree-like Number (as Proportion of Proportion of and branching, and no species is any “higher” than any density/m2) original cohort original cohort other a forms are adapted only to the environments they Age interval surviving to surviving to dying during live in. (days) day x day x interval 4. Darwin’s theory of natural selection and Mendel’s theory of heredity. 0–250 100 100 0.9989 251–500 0.11 0.11 0.273 Chapter 2 501–750 0.08 0.08 0.75 1. The terms are explained in the chapter. 751–1,000 0.06 0.06 – 2. (i) 100% AA; (ii) 1 AA :1 Aa; (iii) 1 AA :2 Aa :1 aa; (iv) 100% AB/AB; and (v) 100% AB/AB. 2. (a) See Section 4.2. (b) In technical terms, drift (on which 3. As fractions: (i) 1/4 AB, 1/4 Ab, 1/4 aB, 1/4 ab; and see Chapter 6). The gene frequencies would change between (ii) (1– r)/2 AB, (1 − r)/2 ab, r/2 Ab, r/2 aB. As ratios: (i) 1 AB : generations because there is heritability (condition 2) and 1 Ab :1 aB :1 ab; and (ii) (1 − r) AB : (1 − r) ab : r Ab : r aB. some individuals produce more offspring than others (condition 3). But if the differences in reproduction are not Chapter 3 systematically associated with some character or other, the 1. Approximately the species level; the pigeon example might changes in gene frequency between generations will be ran- stretch it to genera, but higher categories do not evolve in dom or directionless. (c) No evolution at all. If the character human lifetimes. conferring higher than average fitness is not inherited by the 2. (i) If you look at any one time and place, living things usually individual’s offspring, natural selection cannot increase its fall into distinct, recognizable groups that could be called frequency in the population. “kinds.” (ii) If you look over a range of space (if the “kinds” 3. The requirements of inheritance and association between high in question are species) the kinds break down; if you look reproductive success and some character have also to be met. through a range of times (if the kinds are species or any 4. The mechanism has to: (i) perceive the change in environ- higher category) the kinds also break down. The differences ment; (ii) work out what the appropriate adaptation is to between higher categories can also be broken down by study- the new environment; (iii) alter the genes in the germ line ing the full range of diversity on Earth: you might think that in a manner to code for the new adaptation. (i) is possible; plants and animals are less clear categories after studying the (ii) could vary from possible in a case such as simple range of unicellular organisms. camouflage to impossible in a case requiring a new complex 3. (a), (b), and (d) are homologies; (c) is an analogy. adaptation, such as the adaptations for living on land of the .. Answers to Study and Review Questions 691 first terrestrial tetrapods; and (iii) would contradict what is we do need the second-order assumption that these factors known about genetics and it is difficult to see how it could be are the same for all genotypes.) done. The mechanism would have to work backwards from 6. AA 1/2; aa 1/2. The gene frequencies are 0.5 and the Hardy– the new phenotype (something like a long neck in a giraffe) Weinberg ratio 1/4 : 1/2 : 1/4. The observed to expected to deduce the needed genetic changes, even though the phe- ratios are 2/3 : 4/3 : 2/3, which when scaled to a maximum notype was produced by multiple, interacting genetic and of 1 give the fitnesses 1/2:1:1/2. environmental effects. 7. (a) 0.5; and (b) 0.5625. You need equation 5.13; t = 1. For 5. (a) Directional selection (for smaller brains); (b) stabilizing (a) it looks like this: 0.625 = 0.5 + (0.75 − 0.5)(1 − m). And selection; and (c) no selection. for (b) it looks like this: x = 0.5 + (0.625 − 0.5)(1 − 0.5). 6. (a) Here are two arguments. (i) If every pair produced two 8. (a) The aa genotype is likely to be fixed. If aa mate only offspring, natural selection would favor new genetic variants among themselves and AA and Aa mate only with AA and that produced three, or more, offspring. After the more Aa, whenever there is an Aa × Aa mating, some aa progeny fecund form had spread through the population the average are produced, who will subsequently only mate with other would still be two but the greater competition among indi- aa individuals. (b) Now AA mate only with AA, Aa with Aa, viduals to survive would lead to variation in the success of and aa with aa. The homozygous matings preserve their the broods of different parents. (ii) Random accidents alone genotypes, but when Aa mate together they produce 1/4 aa will guarantee that some individuals fail to breed; then for and 1/4 AA progeny. In an extreme case, the population the average to be two, as it must be for any population that diverges into two species, one AA the other aa and the is reasonably stable in the long term, all successfully repro- heterozygotes are lost. (c) (i) The dominant allele will be ducing individuals will produce more than two offspring. fixed; and (ii) the recessive allele will be fixed. (b) Ecologists discuss this in terms of r and K selection, or life history theory: in some environments there is little competi- ps(1 − ) 9. p′ = tion and selection favors producing large numbers of small 12 −− p2 s pqs offspring; whereas in others there is massive competition and selection favors producing fewer offspring and investing The denominator can be variously rearranged. a large amount in each. Many other factors can also operate. m 10. p* ≈ Chapter 5 s 1. Populations 1 and 5 are in Hardy–Weinberg equilibrium; populations 2–4 are not. As to why they are not, in popula- The derivation starts with the equilibrium condition, tion 4 it looks like AA is lethal, in 2 there may be a hetero- p2s = qm. We then note that q ≈ 1, and p2s ≈ m; divide both zygote advantage, and in 3 a heterozygote disadvantage. sides by s and take square roots. Population 2 could also be produced by disassortative mating, and 3 by assortative mating. In population 3 there Chapter 6 could also be a Wahlund effect. All the deviations are 1. Either 100% A or 100% a (there is an equal chance of each). so large that random sampling is unlikely to be the whole 2. See (a) Section 6.1, and (b) Section 6.3. explanation. 3. (1) 0.5; (2) 0.5; (3) 0.375; and (4) 0. See Section 6.5. 2. (a) p2/(1 − sq2); (b) p/(1 − sq2); and (c) 1 − sq2. 4. (a) and (b) 10−8. Population size cancels out in the formula 3. (1/3)(3 − s) = 1 − (s/3). for the rate of neutral evolution. 4. If you do it in your head, s ≈ 0.1. To be exact, 5. (a) 1/(2N); and (b) (1 − (1/(2N)). s = 0.095181429619. 6. Both manipulations requires substituting 1 − H for f and 5. That the fitness differences are in survivorship (not fertil- then some canceling and multiplying though by −1 to make ity), and in particular in survivorship during the life stage the sign positive. investigated in the mark–recapture experiment. (By the way, mark–recapture experiments are also used by eco- Chapter 7 logists to estimate absolute survival rates: they require the 1. See Figure 7.1a and b. additional assumptions that the animals do not become 2. The main observations suggesting neutral molecular evolu- “trap shy” or “trap happy,” and that the mark and release tion are not also seen in morphology. This was discussed for treatment does not reduce survival. These assumptions are the constancy of evolutionary rates in Section 7.3. The other not needed when estimating relative survival. However, original observations (for absolute rates and heterozygosities, .. 692 Answers to Study and Review Questions and for the relation between rate and constraint) either have 2. Populations 1 and 4 may show fitness epistasis, in which, in not been made, or such observations as there are do not sug- population 1, A1 has higher fitness in combination with B1 gest that the problems found in selective explanations for than with B2, and vice versa in poulation 2.
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